Two dancers Seema and Reema appeared in the audition for a dance competition. The probability that Seema will be selected in this dance competition is 0.05 and that Reema will be selected is 0.10. The probability that both will get selected is 0.02. Determine the probability that: (i) Both Seema and Reema will not get selected for the competition. (ii) At least one of them will not be selected for the competition (iii) Only one of them will be selected for the competition
0.87, 0.98, 0.11 respectively
Probability Calculation for Dance Competition Selection
Let's determine the probabilities for Seema and Reema's selection in the dance competition. We are given the following probabilities regarding their selection status:
- Probability that Seema will be selected, denoted as \(P(S) = 0.05\)
- Probability that Reema will be selected, denoted as \(P(R) = 0.10\)
- Probability that both Seema and Reema will be selected, denoted as \(P(S \cap R) = 0.02\)
Understanding Probability Concepts
To solve this problem, we will use fundamental concepts of probability. These include:
- Union of Events: The probability that at least one of two events, A or B, occurs is given by the formula: \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\).
- Complement of an Event: The probability that an event A does not occur is \(P(A') = 1 - P(A)\).
- De Morgan's Laws: These laws are very useful for expressing the complement of unions and intersections. Specifically, \(P(A' \cap B') = P((A \cup B)')\) and \(P(A' \cup B') = P((A \cap B)')\).
(i) Probability Both Seema and Reema Will Not Get Selected
This part asks for the probability that neither Seema nor Reema is selected. In probability notation, this is \(P(S' \cap R')\), where \(S'\) means Seema is not selected and \(R'\) means Reema is not selected.
According to De Morgan's Law, \(P(S' \cap R')\) is equivalent to \(P((S \cup R)')\). This means the probability that both are not selected is the complement of the probability that at least one of them is selected.
First, we calculate the probability that at least one of them is selected, \(P(S \cup R)\):
\[P(S \cup R) = P(S) + P(R) - P(S \cap R)\]Substitute the given values:
\[P(S \cup R) = 0.05 + 0.10 - 0.02\] \[P(S \cup R) = 0.15 - 0.02\] \[P(S \cup R) = 0.13\]Now, we find the probability that both are not selected by taking the complement of \(P(S \cup R)\):
\[P(\text{Both not selected}) = 1 - P(S \cup R)\] \[P(\text{Both not selected}) = 1 - 0.13\] \[P(\text{Both not selected}) = 0.87\](ii) Probability At Least One of Them Will Not Be Selected
This asks for the probability that Seema is not selected, or Reema is not selected, or both are not selected. In probability notation, this is \(P(S' \cup R')\).
Using De Morgan's Law, \(P(S' \cup R')\) is equivalent to \(P((S \cap R)')\). This means the probability that at least one of them is not selected is the complement of the probability that both of them ARE selected.
We are already given the probability that both Seema and Reema will be selected, which is \(P(S \cap R) = 0.02\).
So, the probability that at least one of them will not be selected is:
\[P(\text{At least one not selected}) = 1 - P(S \cap R)\] \[P(\text{At least one not selected}) = 1 - 0.02\] \[P(\text{At least one not selected}) = 0.98\](iii) Probability Only One of Them Will Be Selected
This scenario means one of two mutually exclusive events occurs: either Seema is selected AND Reema is not selected, OR Reema is selected AND Seema is not selected. We can sum their individual probabilities.
First, calculate the probability that Seema is selected and Reema is not selected, \(P(S \cap R')\):
\[P(S \cap R') = P(S) - P(S \cap R)\]Substitute the given values:
\[P(S \cap R') = 0.05 - 0.02\] \[P(S \cap R') = 0.03\]Next, calculate the probability that Reema is selected and Seema is not selected, \(P(R \cap S')\):
\[P(R \cap S') = P(R) - P(S \cap R)\]Substitute the given values:
\[P(R \cap S') = 0.10 - 0.02\] \[P(R \cap S') = 0.08\]Finally, the probability that only one of them will be selected is the sum of these two probabilities:
\[P(\text{Only one selected}) = P(S \cap R') + P(R \cap S')\] \[P(\text{Only one selected}) = 0.03 + 0.08\] \[P(\text{Only one selected}) = 0.11\]Summary of Probabilities
Here is a summary of the calculated probabilities for the dance competition selection:
| Scenario | Calculated Probability |
|---|---|
| (i) Both Seema and Reema will not get selected | 0.87 |
| (ii) At least one of them will not be selected | 0.98 |
| (iii) Only one of them will be selected | 0.11 |
The calculated probabilities are 0.87, 0.98, and 0.11 respectively.
Important Questions from Conditional Probability
Two events A and B are such that P(not B) = 0.8, P(A ∪ B) = 0.5 and P(A|B) = 0.4. Then P(A) is equal to
For two events, A and B, it is given that \({\rm{P}}\left( {\rm{A}} \right) = \frac{3}{5},{\rm{\;P}}\left( {\rm{B}} \right) = \frac{3}{{10}}\) and \({\rm{P}}\left( {{\rm{A|B}}} \right) = \frac{2}{3}\) . If A̅ and B̅ are the complementary events of A and B, then what is P(A̅ | B̅) equal to?
For two mutually exclusive events A and B, P(A) = 0.2 and P (A̅ ∩ B) = 0.3. What is P (A|(A ∪ B)) equal to?
If an event B has occurred and has P(B) = 1, the conditional probability P(A|B) is equal to:
If P(A) = 0.7, P(B) = 0.5 and P(B/A) = 0.3, find (i) P(A/B) (ii) P(A ∪ B)?
